7.2 Q-17 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 17

Problem Statement

Using the Binomial Theorem:

(a) Find the remainder when $7^{100}$ is divided by $6$ .

(b) Find the last digit (units digit) of $3^{100}$ .

Part (a): Remainder when $7^{100}$ is divided by $6$

Strategy: Write $7$ as $(6 + 1)$ so that the Binomial Theorem produces terms that are multiples of $6$ , leaving a simple remainder.

$7^{100} = (6 + 1)^{100}$

Expanding by the Binomial Theorem:

$= (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6^{1} + (100 100) 6^{0}$

$= (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6 + 1$

Every term except the last contains a factor of $6$ , so:

$7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Part (b): Last digit of $3^{100}$

Strategy: Write $3^{100} = (3^{2})^{50} = 9^{50}$ . Then write $9 = (10 - 1)$ and expand.

$9^{50} = (10 - 1)^{50}$

Expanding by the Binomial Theorem:

$= (0 50) 1 0^{50} - (1 50) 1 0^{49} + \dots - (49 50) 1 0^{1} + (50 50) (- 1)^{50}$

Every term except the last is a multiple of $10$ , so:

$9^{50} = 10 m + (- 1)^{50} = 10 m + 1 for some integer m$

The units digit is determined by the last term:

$Last digit of 3^{100} = 1$

Key Technique Summary

Goal	Strategy
Find remainder when $a^{n} \div d$	Write $a = (d \cdot k \pm 1)$ , expand, all terms except $\pm 1$ are divisible by $d$
Find last digit of $a^{n}$	Write $a^{n}$ in terms of $(10 \pm r)^{m}$ , expand; last digit comes from the constant term

Exercise 7.2 — Question 17

Problem Statement

Using the Binomial Theorem:

(a) Find the remainder when $7^{100}$ is divided by $6$ .

(b) Find the last digit (units digit) of $3^{100}$ .

Part (a): Remainder when $7^{100}$ is divided by $6$

Strategy: Write $7$ as $(6 + 1)$ so that the Binomial Theorem produces terms that are multiples of $6$ , leaving a simple remainder.

$7^{100} = (6 + 1)^{100}$

Expanding by the Binomial Theorem:

$= (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6^{1} + (100 100) 6^{0}$

$= (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6 + 1$

Every term except the last contains a factor of $6$ , so:

$7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Part (b): Last digit of $3^{100}$

Strategy: Write $3^{100} = (3^{2})^{50} = 9^{50}$ . Then write $9 = (10 - 1)$ and expand.

$9^{50} = (10 - 1)^{50}$

Expanding by the Binomial Theorem:

$= (0 50) 1 0^{50} - (1 50) 1 0^{49} + \dots - (49 50) 1 0^{1} + (50 50) (- 1)^{50}$

Every term except the last is a multiple of $10$ , so:

$9^{50} = 10 m + (- 1)^{50} = 10 m + 1 for some integer m$

The units digit is determined by the last term:

$Last digit of 3^{100} = 1$

Key Technique Summary

Goal	Strategy
Find remainder when $a^{n} \div d$	Write $a = (d \cdot k \pm 1)$ , expand, all terms except $\pm 1$ are divisible by $d$
Find last digit of $a^{n}$	Write $a^{n}$ in terms of $(10 \pm r)^{m}$ , expand; last digit comes from the constant term